Q16E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution for $\mathbf{t}\mathbf{<}\mathbf{0}$.

${\mathbf{t}}^{\mathbf{2}}\mathbf{y}\text{'}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{3}\mathbf{ty}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The general solution of the given equation ${\mathrm{t}}^2\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)-3\mathrm{ty}\text{'}\left(\mathrm{t}\right)+6\mathrm{y}\left(\mathrm{t}\right)=0$ is:

$\mathrm{y}={\mathrm{c}}_1{\mathrm{t}}^2\cos (\sqrt{2}\mathrm{lnt})+{\mathrm{c}}_2{\mathrm{t}}^2\sin (\sqrt{2}\mathrm{lnt})$ for $\mathrm{t}<0$.

Step 1: Substitute the values

Given differential equation is ${\mathrm{t}}^2\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)-3\mathrm{ty}\text{'}\left(\mathrm{t}\right)+6\mathrm{y}\left(\mathrm{t}\right)=0$

 

Assume $\mathrm{y}={\mathrm{t}}^{\mathrm{r}}$, then $\mathrm{y}\text{'}={\mathrm{rt}}^{\mathrm{r}-1}$

$\mathrm{y}\text{'}\text{'}=\mathrm{r}(\mathrm{r}-1){\mathrm{t}}^{\mathrm{r}-2}$

 

Substitute these equations in the differential equation

$\begin{array}{rcl}{\mathrm{t}}^2\mathrm{r}(\mathrm{r}-1){\mathrm{t}}^{\mathrm{r}-2}-3{\mathrm{trt}}^{\mathrm{r}-1}+6{\mathrm{t}}^{\mathrm{r}}& =& 0\\ \left(\mathrm{r}\right(\mathrm{r}-1)-3\mathrm{r}+6){\mathrm{t}}^{\mathrm{t}}& =& 0\\ {\mathrm{r}}^2-4\mathrm{r}+6& =& 0\end{array}$

Step 2: Finding roots

Find the roots of this equation

$$\begin{gathered} \mathrm{r}=\frac{4\pm \sqrt{4^2-4\times 6\times 1}}{2\times 1} \\ \mathrm{r}=\frac{4\pm \sqrt{16-24}}{2} \\ \mathrm{r}=\frac{4\pm \sqrt{-8}}{2} \\ \mathrm{r}=\frac{4\pm 2\sqrt{2}\mathrm{i}}{2} \\ \mathrm{r}=2\pm \sqrt{2}\mathrm{i} \end{gathered}$$ 

Therefore, the general solution is $\mathrm{y}={\mathrm{c}}_1{\mathrm{t}}^2\cos (\sqrt{2}\mathrm{lnt})+{\mathrm{c}}_2{\mathrm{t}}^2\sin (\sqrt{2}\mathrm{lnt})$.